Number of distinct real roots of the equation $x^{4}+4x^{3}-2x^{2}-12x+k=0$ is

We can factor the quartic expression x^4 + 4x^3 - 2x^2 - 12x + k = 0 into a product of two quadratics, such as (x^2 + 2x)^2 - 6(x^2 + 2x) + k = 0. Letting y = x^2 + 2x, the equation transforms into the quadratic y^2 - 6y + k = 0. The discriminant of this quadratic in y is (-6)^2 - 4(1)(k) = 36 - 4k. If k < -7, the discriminant 36 - 4k is strictly positive, resulting in two distinct real roots for y. Because k is negative, the product of the roots of the quadratic in y is negative, forcing one positive and one negative y-value, and solving x^2 + 2x = y for a negative y always yields two distinct real roots for x. Consequently, the original quartic equation will have exactly two distinct real roots.

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